{"id":8498,"date":"2024-09-18T11:16:01","date_gmt":"2024-09-18T05:46:01","guid":{"rendered":"https:\/\/samacheerkalvi.guide\/?p=8498"},"modified":"2024-09-19T10:00:16","modified_gmt":"2024-09-19T04:30:16","slug":"samacheer-kalvi-8th-maths-guide-chapter-5-ex-5-3","status":"publish","type":"post","link":"https:\/\/samacheerkalvi.guide\/samacheer-kalvi-8th-maths-guide-chapter-5-ex-5-3\/","title":{"rendered":"Samacheer Kalvi 8th Maths Guide Chapter 5 Geometry Ex 5.3"},"content":{"rendered":"

Tamilnadu State Board New Syllabus Samacheer Kalvi 8th Maths Guide<\/a> Pdf Chapter 5 Geometry Ex 5.3 Text Book Back Questions and Answers, Notes.<\/p>\n

Tamilnadu Samacheer Kalvi 8th Maths Solutions Chapter 5 Geometry Ex 5.3<\/h2>\n

Question 1.
\nIn the figure, given that \u22201 = \u22202 and \u22203 \u2261 \u22204. Prove that \u2206 MUG \u2261 \u2206TUB.
\n\"Samacheer
\nAnswer:<\/p>\n\n\n\n\n\n\n\n
Statements<\/td>\nReasons<\/td>\n<\/tr>\n
1. In \u25b3MUG and \u25b3TUG<\/p>\n

Mu = TU<\/td>\n

\u22203 = \u22204, opposite sides of equal angles<\/td>\n<\/tr>\n
2. UG = UB<\/td>\n\u22201 = \u22202<\/p>\n

Side opposite to equal angles are equal<\/td>\n<\/tr>\n

3. \u2220GUM = \u2220BUT<\/td>\nVertically opposite angle<\/td>\n<\/tr>\n
4. \u2206MUG \u2261 \u00a0\u2206TUG<\/td>\nSAS criteria<\/p>\n

By 1,2 and 3<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

\"Samacheer<\/p>\n

Question 2.
\nFrom the figure, prove that \u2206SUN ~ \u2206RAY.
\n\"Samacheer
\nAnswer:
\nProof: from the \u2206 SUN and \u2206RAY
\nSU = 10
\nUN = 12
\nSN = 14
\nRA = 5
\nAY = 6
\nRY = 7
\n\"Samacheer
\nFrom (1), (2) and (3) we have
\n\"Samacheer
\nThe sides are proportional
\n\u2234 \u2206SUN ~ \u2206RAY<\/p>\n

\"Samacheer<\/p>\n

Question 3.
\nThe height of a tower is measured by a mirror on the ground at R by which the top of the tower\u2019s reflection is seen. Find the height of the tower. If \u2206PQR ~ \u2206STR
\n\"Samacheer
\nAnswer:
\nThe image and its reflection make similar shapes
\n\u2234 \u2206PQR ~ \u2206STR
\n\"Samacheer
\n\u21d2 \\(\\frac{h}{8}=\\frac{60}{10}\\)
\nh = \\(\\frac{60}{10}\\) \u00d7 8
\n= 48 feet
\n\u2234 Height of the tower = 48 feet.<\/p>\n

\"Samacheer<\/p>\n

Question 4.
\nFind the length of the support cable required to support the tower with the floor.
\n\"Samacheer
\nAnswer:
\nFrom the figure, by Pythagoras theorem,
\nx2<\/sup> = 202<\/sup> + 152<\/sup>
\n= 400 + 225 = 625
\nx2<\/sup> = 252<\/sup> \u21d2 x = 25ft.
\n\u2234 The length of the support cable required to support the tower with the floor is 25ft.<\/p>\n

\"Samacheer<\/p>\n

Question 5.
\nRithika buys an LED TV which has a 25 inches screen. If its height is 7 inches, how wide is the screen? Her TV cabinet is 20 inches wide. Will the TV fit into the cabinet? Give reason.
\nAnswer:
\n\"Samacheer
\nTake the sides of a right angled triangle \u2206ABC as
\na = 7 inches
\nb = 25 inches
\nc = ?
\nBy Pythagoras theorem,
\nb2<\/sup> = a2<\/sup> + c2<\/sup>
\n252<\/sup> = 72<\/sup> + c2<\/sup>
\n\u21d2 c2<\/sup> = 252<\/sup> – 72<\/sup> = 625 – 49 = 576
\n\u2234 c2<\/sup> = 242<\/sup>
\n\u21d2 c = 24 inches
\n\u2234 Width of TV cabinet is 20 inches which is lesser than the width of the screen ie.24 inches.
\n\u2234 The TV will not fit into the cabinet.<\/p>\n

\"Samacheer<\/p>\n

Challenging Problems<\/p>\n

Question 6.
\nIn the figure, \u2220TMA \u2261\u2220IAM and \u2220TAM \u2261 \u2220IMA. P is the midpoint of MI and N is the midpoint of AI. Prove that \u0394 PIN ~ \u0394 ATM.
\n\"Samacheer
\nAnswer:
\nproof:
\n\"Samacheer<\/p>\n

\"Samacheer<\/p>\n

Question 7.
\nIn the figure, if \u2220FEG \u2261 \u22201 then, prove that DG2<\/sup> = DE.DF.
\n\"Samacheer
\nAnswer:
\nProof:
\n\"Samacheer<\/p>\n

\"Samacheer<\/p>\n

Question 8.
\nThe diagonals of the rhombus is 12 cm and 16 cm. Find its perimeter. (Hint: the diagonals of rhombus bisect each other at right angles).
\nAnswer:
\n\"Samacheer
\nHere AO = CO = 8cm
\nBO = DO = 6cm
\n(\u2234the diagonals of rhombus bisect each other at right angles)
\n\u2234 In \u2206 AOB, AB2<\/sup> = AO2<\/sup> + OB2<\/sup>
\n= 82<\/sup> + 62<\/sup> = 64 + 36
\n= 100 = 102<\/sup>
\n\u2234 AB = 10
\nSince it is a rhombus, all the four sides are equal.
\nAB = BC = CD = DA
\n\u2234 Its Perimeter = 10 + 10 + 10 + 10 = 40 cm<\/p>\n

\"Samacheer<\/p>\n

Question 9.
\nIn the figure, find AR.
\n\"Samacheer
\nAnswer:
\n\u2206 AFI, \u2206 FRI are right triangles.
\nBy Pythagoras theorem,
\nAF2<\/sup> = AI2<\/sup> – FI2<\/sup>
\n= 252<\/sup> – 152<\/sup>
\n= 625 – 225 = 400 = 202<\/sup>
\n\u2234 AF = 20ft.
\nFR2<\/sup> = RI2<\/sup> – FI2<\/sup>
\n= 172<\/sup> – 152<\/sup> = 289 – 225 = 64 = 82<\/sup>
\nFR = 8ft.
\n\u2234 AR = AF + FR
\n= 20 + 8 = 28 ft.<\/p>\n

\"Samacheer<\/p>\n

Question 10.
\nIn \u2206DEF, DN, EO, FM are medians and point P is the centroid. Find the following.
\n(i) IF DE = 44, then DM = ?
\n(ii) IFPD=12, then PN= ?
\n(iii) IfDO = 8, then PD = ?
\n(iv) IF 0E = 36 then EP = ?
\n\"Samacheer
\nAnswer:
\nGiven DN, EO, FM are medians.
\n\u2234 FN = EN
\nDO = FO
\nEM = DM<\/p>\n

(i) If DE = 44,then
\nDM = \\(\\frac{44}{2}\\) = 22
\nDM = 22<\/p>\n

(ii) If PD = 12,PN = ?
\n\\(\\frac{P D}{P N}=\\frac{2}{1}\\)
\n\\(\\frac{12}{\\mathrm{PN}}=\\frac{2}{1}\\) \u21d2 PN = \\(\\frac{12}{2}\\) = 6.
\nPN = 6<\/p>\n

\"Samacheer<\/p>\n

(iii) If DO = 8, then
\nFD = DO + OF
\n= 8 + 8
\nFD = 16<\/p>\n

(iv) If OE = 36
\n\"Samacheer
\nPE = 24<\/p>\n","protected":false},"excerpt":{"rendered":"

Tamilnadu State Board New Syllabus Samacheer Kalvi 8th Maths Guide Pdf Chapter 5 Geometry Ex 5.3 Text Book Back Questions and Answers, Notes. Tamilnadu Samacheer Kalvi 8th Maths Solutions Chapter 5 Geometry Ex 5.3 Question 1. In the figure, given that \u22201 = \u22202 and \u22203 \u2261 \u22204. Prove that \u2206 MUG \u2261 \u2206TUB. Answer: …<\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_jetpack_memberships_contains_paid_content":false,"footnotes":"","jetpack_publicize_message":"","jetpack_publicize_feature_enabled":true,"jetpack_social_post_already_shared":false,"jetpack_social_options":{"image_generator_settings":{"template":"highway","enabled":false},"version":2}},"categories":[10],"tags":[],"class_list":["post-8498","post","type-post","status-publish","format-standard","hentry","category-class-8"],"jetpack_publicize_connections":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/posts\/8498"}],"collection":[{"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/comments?post=8498"}],"version-history":[{"count":1,"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/posts\/8498\/revisions"}],"predecessor-version":[{"id":40867,"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/posts\/8498\/revisions\/40867"}],"wp:attachment":[{"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/media?parent=8498"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/categories?post=8498"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/tags?post=8498"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}